Answer:
a) 358.8 KJ/kg
b) 0.0977 KJ/K- kg
c) 83.28%
Explanation:
N2 at 300 k. ( use the properties of N2 at 300 k (T1) )
Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk , R = 0.1297 KJ/kgk , y = 1.4 ,
Given data:
T2 = 645 k
P1 = 1 bar , P2 = 10.5 bar
<u>a)Determine the work input in KJ/Kg of N2 flowing </u>
Winput = h2 - h1 = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg
<u>b) Determine the rate of entropy in KJ/K- kg of N2 flowing </u>
Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1
= 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )
= 0.0977 KJ/K- kg
<u>c) Determine isentropic compressor efficiency </u>
Isentropic compressor efficiency = 83.28%
calculated using the relation below
( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )
where T'2 = 587.314
Answer:
Questions are answered in the explanation section.
Explanation:
a) Of both metals, lead will corrode, since it reacts with seawater to form oxides, chlorides, etc., while bronze is more resistant to oxidation.
b) The corrosion rate will remain constant, because the concentration of seawater does not change over time, therefore, the corrosion rate is independent of the area that is exposed to seawater.
c) Corrosion can be reduced by using protective coatings, so that the metal is isolated from sea water, for example, paints, acrylic powders, etc.
Answer:
Base temperature is 46.23 °C
Explanation:
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